Question

How can three resistors of resistance 2ohm,3ohm and 6ohm be connected to go live total resistance of (a) 4ohm,(b)1ohm?​

Answer 1

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(a) R2, R3 are parallel and series with R1

(b) R1, R2 and R3 are in parallel

Step-by-step explanation:

`\sf R_1 = 2 \: ohm\\ \sf R_2 = 3 \: ohm \\ \sf R_3 = 6 \: ohm`

(a)

`\sf R_2 \: and \: R_3 \: are \: in \: parallel \\ \sf so \: let \: the \: total \: of \: R_2 \: and \: R_3 \: be \: R_a`

`\sf \large (1)/(R_a) = (1)/(R_2) + (1)/(R_3) \: as \: they \: are \: in \: parallel`

`\sf R_a = (1)/(3) + (1)/(6) \\ \\ \sf (1)/(R_a) = (2 + 1)/(6) \\ \\ \sf (1)/(R_a) = (3)/(6) \\ \\ \sf (1)/(R_a) = (1)/(2) \\ \\ \sf R_a = 2 \: ohm`

Ra and R1 is in series

And there total will be R

`\sf R = R_a + R_1 \\ \\ \sf R = 2 + 2 \\ \\ \sf R = 4 \: ohm`

(b)

`\sf R_1 R_2 \: and \:R_3 \: are \: in \: parallel \: so \: total \: be \: R`

`\sf (1)/(R) = (1)/(R_1) + (1)/(R_2) + (1)/(R_3) \\ \\ \sf (1)/(R) = (1)/(2) + (1)/(3) + (1)/( 6) \\ \\ \sf (1)/(R) = (3+ 2 +1 )/(6) \\ \\ \sf (1)/(R) = (6)/(6) \\ \\ \sf R = 1 \: ohm`